A perpendicular to the line y−4x+3=0 passes through the point (-8, 1/3). Determine its eqn. Find the equation of a line which is equidistant from the points (2,3) and (6,1), expressing in the form 9x+by=c where a,b and c are constants.

Mlinganyo uliotolewa ni $y - 4x + 3 = 0$. Tunaweza kuandika upya kama $y = 4x - 3$.
Gradient ya mstari huu, $m_1$, ni 4.

Kwa mistari miwili iliyo wima, bidhaa ya gradient zao ni $-1$.

Gradient ya mstari ulio wima ni $-\frac{1}{4}$.

Mstari ulio wima unapita kwenye nukta $(-8, 5)$ na una gradient $m_2 = -\frac{1}{4}$. Tunatumia fomu $y - y_1 = m(x - x_1)$.

Zidisha pande zote mbili kwa 4:

Panga upya ili kupata fomu $ax + by = c$:

Mlinganyo wa mstari ni $\boxed{\mathbf{x + 4y = 12}}$.

Nukta ya kati $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

Gradient $m_1 = \frac{y_2 - y_1}{x_2 - x_1}$.

Mstari unaotafutwa ni perpendicular bisector (mstari unaogawanya sehemu ya mstari kwa nusu na kuwa wima kwake). Gradient yake, $m_2$, itakuwa:

Mstari unapita kwenye nukta $(4, 2)$ na una gradient $m_2 = 2$. Tunatumia fomu $y - y_1 = m(x - x_1)$.

Panga upya:

Mlinganyo wa mstari ni $\boxed{\mathbf{2x - y = 6}}$.